Question

Calculate the derivative of the following functions. (a) f(x)=(ln(cosx))5

Answer 1

Explanation:

To find the derivative of the function f(x) = (ln(cos(x)))^5, we can use the chain rule.

Let's break it down step by step:

Step 1: Identify the inner function and let u = ln(cos(x)).

- u = ln(cos(x))

Step 2: Differentiate the inner function u = ln(cos(x)).

- du/dx = d/dx [ln(cos(x))]

Using the chain rule, the derivative of ln(u) is given by (1/u) * du/dx.

- du/dx = (1/cos(x)) * (-sin(x)) = -sin(x)/cos(x) = -tan(x)

Step 3: Substitute the inner function back into the original function.

- f(x) = u^5 = (ln(cos(x)))^5

Step 4: Apply the chain rule to find the derivative of the original function f(x).

- df/dx = d/dx[(ln(cos(x)))^5]

= 5(ln(cos(x)))^4 * d/dx[ln(cos(x))]

= 5(ln(cos(x)))^4 * (-tan(x))

= -5(ln(cos(x)))^4 * tan(x)

Therefore, the derivative of f(x) = (ln(cos(x)))^5 is -5(ln(cos(x)))^4 * tan(x).

Answer 2

`f'(x)=-5(\ln(\cos x))^4\tan x`

Explanation:

To calculate the derivative of the given function f(x) = (ln(cos(x)))⁵, use the chain rule for differentiation.

The chain rule is used for differentiating functions of functions.

`\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}*\frac{\text{d}u}{\text{d}x}$\\\end{minipage}}`

Given function:

`f(x) = (\ln(\cos x))^5`

Begin by deciding which part of the function to replace with u.

In this case, let u = ln(cos x). Therefore:

`y = (\ln(\cos x))^5\;\;\textsf{can be written as}\;\;y = u^5,\;\;\textsf{where}\;\;u = \ln(\cos x).`

To find dy/dx, we need to differentiate y with respect to u, and differentiate u with respect to x.

Differentiate y with respect to u:

`\frac{\text{d}y}{\text{d}u}=5u^4=5(\ln(\cos x))^4`

To differentiate u with respect to x, we also need to use the chain rule.

For u = ln(cos x), the bit inside the brackets is easy to differentiate, so let v = cos x. Therefore:

`u=\ln(\cos x)\;\;\textsf{can be written as}\;\;u=\ln (v),\;\;\textsf{where}\;\;v=\cos x`

`\textsf{So},\;\;\frac{\text{d}u}{\text{d}v}=(1)/(v)=(1)/(\cos x), \;\; \textsf{and}\;\;\frac{\text{d}v}{\text{d}x}=-\sin x`

Put everything into the chain rule to find du/dx:

`\frac{\text{d}u}{\text{d}x}=\frac{\text{d}u}{\text{d}v}* \frac{\text{d}v}{\text{d}x}`

`\frac{\text{d}u}{\text{d}x}=(1)/(\cos x)*-\sin x`

`\frac{\text{d}u}{\text{d}x}=-(\sin x)/(\cos x)`

`\frac{\text{d}u}{\text{d}x}=-\tan x`

Now we have found dy/du and du/dx, we can substitute them into the chain rule to find dy/dx:

`\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}* \frac{\text{d}u}{\text{d}x}`

`\frac{\text{d}y}{\text{d}x}=5(\ln(\cos x))^4 * (-\tan x)`

`\frac{\text{d}y}{\text{d}x}=-5(\ln(\cos x))^4\tan x`

Therefore, the derivative of the given function f(x) is:

`\large\boxed{f'(x)=-5(\ln(\cos x))^4\tan x}`